题目

已知A（8,0）,B、C两点分别在y轴和x轴上运动,并且满足向量AB．向量BP=0,向量BC=向量CP；（1）求动点P的轨迹方程；（2）若过点A的直线l与动点P的轨迹交于M、N两点,且向量QM．向量QN=97,其中Q（-1,0）,求直线l的方程?详细过程,第一问的答案是y^2=-4x 只要第二问的过程

提问时间：2021-12-20

答案

(1)
let B(0,y1),C(x1,0)
let P(x,y)
BC=CP
=>(x1,-y1) = (x-x1,y)
=> x1=x/2 and y1= -y
AB.BP=0
(8,y1).( x,y-y1) =0
(8,-y).(x,2y)=0
8x-2y^2 =0
locus of P 4x-y^2 =0
(2)
let the line(l) be y = mx+c,
l pass through (8,0)
=> 8m+c = 0
c=-8m
l:y= mx -8m
let M(x1,y1),N(x2,y2)
lin(l) intersact 4x-y^2 =0 at M,N
=>
4x-(mx-8m)^2 =0
m^2x^2 -(16m^2+4)x +64m^2 =0
x1+x2 = (16m^2+4)/m^2 (1)
x1x2=64 (2)
Similarly
4(y+8m)/m -y^2 =0
my^2-4y-32m=0
y1y2 = -32 (3)
QM.QN=97,Q(-1,0)
=>(x1+1,y1).(x2+1,y2) =97
(x1+1)(x2+1)+y1y2 =97
x1x2+(x1+x2)+y1y2 = 96
64+(16m^2+4)/m^2-32=96
48m^2-4=0
m^2 = 1/12
m = √12/12 or -√12/12
m= √12/12 ,c=-2√12/3
m=- √12/12,c= 2√12/3
the line
y = (√12/12)x -2√12/3 or
y = -(√12/12)x +2√12/3